Speed of sound — a = √(γRT) — derivation: Worked examples

This page is the "no surprises" drill for the speed-of-sound formula. We take $a = γ R T$ and push it through every kind of case an exam or a real wing can throw at you: normal air, cold air, other gases, degenerate inputs (what happens as $T \to 0$ ?), the classic Newton mistake, ratios, and a word problem. Before each example you make a Forecast — guess the answer's ballpark first, then check yourself.

The scenario matrix

Every problem about $a = γ R T$ falls into one of these cells. The worked examples below each carry a tag telling you which cell they nail.

#	Case class	What changes	Example
A	Baseline — standard sea-level air	nothing (reference case)	Ex 1
B	Cold input — lower $T$	$T ↓\Rightarrow a ↓$	Ex 2
C	Different gas — new $γ, R, M$	swap all three constants	Ex 3
D	Ratio / relative — compare two states	constants cancel, only $T$ ratio survives	Ex 4
E	Degenerate / limiting — $T \to 0$ , $T \to \infty$	boundary behaviour	Ex 5
F	Wrong process trap — isothermal vs adiabatic	the $γ$ / Newton error	Ex 6
G	Unit trap — Celsius, or universal $R$	catch the classic slip	Ex 7
H	Real-world word problem — Mach + altitude	chain $a$ into $M$	Ex 8
I	Exam twist — solve backwards for $T$	invert the formula	Ex 9

Worked examples

Worked example Example 5 — Degenerate & limiting inputs

(cell E) What is $a$ as $T \to 0 K$ ? And what does $T \to \infty$ imply? Use air constants where numbers help.

Forecast: at absolute zero molecules stop moving, so nothing carries a push — guess $a \to 0$ . At huge $T$ , $a$ grows without bound (but slowly, like $T$ ).

Limit $T \to 0$ : $a = γ R \cdot 0 = 0$ . Why this step? The product under the root vanishes; a zero factor kills it. Physically: no thermal molecular motion → no carrier for the wave.
Limit $T \to \infty$ : $a = γ R T \to \infty$ , but it rises like $T$ , so slowly — doubling $T$ multiplies $a$ by only $2 \approx 1.41$ . Why this step? The exponent $\frac{1}{2}$ controls how fast a limit blows up; it's sub-linear, so no runaway.
Sanity numbers: at $T = 1152 K$ (= $4 \times 288$ ), $a = 1.4 \cdot 287 \cdot 1152 = 462, 872 \approx 680 m/s$ — exactly $2 \times 340$ , since $4 = 2$ .

Verify: the $T = 0$ answer is $0$ ✓ (no negative or complex speed — always physical because $T \geq 0$ in kelvin). The $4 \times T$ giving $2 \times a$ confirms the $T$ scaling. ✓

Figure — Speed of sound — a = √(γRT) — derivation

Worked example Example 6 — The wrong-process trap: Newton vs Laplace

(cell F) Compute the isothermal (Newton) speed $a_{iso} = R T$ and the correct adiabatic (Laplace) speed $a = γ R T$ at $T = 288 K$ . By what percent is Newton wrong?

Forecast: Newton dropped the $γ$ , so his number is smaller by a factor $γ = 1.4 \approx 1.18$ . Expect ~15% too low.

Newton: $a_{iso} = 287 \times 288 = 82, 656 \approx 287.5 m/s$ . Why this step? Isothermal $p = ρR T$ gives $d p / d ρ = R T$ , no $γ$ — this is exactly the tempting error from the parent note.
Laplace (correct): $a = 1.4 \times a_{iso} = 1.183 \times 287.5 \approx 340 m/s$ . Why this step? The real wave is adiabatic, multiplying $a^{2}$ by $γ$ , i.e. $a$ by $γ$ .
Error: $\frac{340 - 287.5}{340} = 0.154 = 15.4%$ too low.

Verify: the two speeds differ by exactly $1.4 = 1.183$ ✓, and $287.5$ m/s is Newton's famous ~15% miss. ✓

Worked example Example 7 — Unit traps: Celsius & universal

$R$ (cell G) A student writes $a = 1.4 \times 8.314 \times 15$ for "air at $1 5^{\circ}$ C". What two mistakes are there, and what is the correct value?

Forecast: using $R_{u} = 8.314$ instead of $287$ makes the number ~ $287/8.314 \approx 5.9 \times$ too small, and $15$ K instead of $288$ K makes it wildly wrong too. Their answer will be nonsense (tens of m/s).

Their (wrong) value: $1.4 \times 8.314 \times 15 = 174.6 \approx 13.2 m/s$ . Why this step? Show how far off the double error lands — nowhere near a real sound speed.
Fix the temperature: $1 5^{\circ} C = 15 + 273 = 288 K$ . Why this step? The formula needs absolute temperature; Celsius has a false zero.
Fix the constant: use $R = 287$ (specific), not $R_{u} = 8.314$ (universal). Why this step? $a$ is per-kilogram physics; $R_{u}$ is per-mole. Divide by molar mass first: $8314/29 \approx 287$ .
Correct value: $1.4 \times 287 \times 288 \approx 340 m/s$ .

Verify: the wrong answer ( $13.2$ ) is absurd (a bicycle is faster) ✓; the corrected one is $340$ ✓.

Worked example Example 8 — Real-world word problem: Mach at altitude

(cell H) A jet cruises at a true airspeed $V = 250 m/s$ at 11 km where $T = 217 K$ . Is it in the compressible regime? See Compressibility and why M > 0.3 matters.

Forecast: sound at altitude is slow (~295 from Ex 2), so $M = 250/295$ is comfortably above 0.3 — yes, compressible; probably high-subsonic ~0.85.

Local sound speed: $a = 1.4 \times 287 \times 217 \approx 295 m/s$ (from Ex 2). Why this step? Mach uses the local $a$ , set by the local temperature — not the sea-level value.
Mach number: $M = \frac{V}{a} = \frac{250}{295} \approx 0.847$ . Why this step? $M = V / a$ compares the plane's speed to how fast pressure signals travel; it decides whether density changes matter.
Compare to thresholds: $0.847 > 0.3$ → compressible, and $< 1$ → still subsonic.

Verify: $M \approx 0.85$ ✓ — a realistic airliner cruise Mach. Note the same $250$ m/s at sea level ( $a = 340$ ) gives $M = 0.74$ : at altitude the same speed is "more transonic" because $a$ dropped. ✓

Quick recall

Recall Cover the answers — did you hit every cell?

How does $a$ change if only $T$ falls? ::: It drops, as $a \propto T$ (cell B/D).
Ratio of sound speeds at $T_{2}$ and $T_{1}$ ? ::: $T_{2} / T_{1}$ — constants cancel (cell D).
What is $a$ as $T \to 0 K$ ? ::: $0$ — no thermal motion to carry the wave (cell E).
Newton's isothermal error is how large? ::: ~15% too low; he dropped the $γ$ factor (cell F).
Two classic unit slips? ::: Celsius instead of kelvin; universal $R_{u}$ instead of specific $R$ (cell G).
To get $T$ from a measured $a$ ? ::: $T = a^{2} / (γ R)$ (cell I).